Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for (int i = (a), i##_end_ = (b); i < i##_end_; ++i)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

typedef long long LL;

const int oo = 0x3f3f3f3f;

const int maxn = 200100;

int n, m;
int a[maxn + 5];

int col[maxn + 5];

set<pair<int, int> > all;

int ri[maxn + 5];
int pre[maxn + 5];

int fa[maxn + 5];

int main()
{
#ifdef matthew99
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	scanf("%d", &n);
	REP(i, 0, n) scanf("%d", a + i), --a[i];
	all.insert(mp(-1, 0));
	m = 0;
	REP(i, 0, n)
	{
		auto u = --all.lower_bound(mp(a[i], 0));
		pair<int, int> to_add = mp(a[i], u->y);
		int nxt = -1;
		if (!~u->x) nxt = u->y + 1, ++m;
		ri[u->y] = a[i];
		pre[a[i]] = u->x;
		col[a[i]] = u->y;
		all.erase(u);
		all.insert(to_add);
		if (~nxt) all.insert(mp(-1, nxt));
	}
	queue<int> q;
	REP(i, 0, m) q.push(i);
	int ans = 0;
	int lst = n - 1;
	vector<int> seq;
	for (int i = n - 2; i >= 0; --i)
	{
		if (col[i] != col[i + 1] && (col[i] - col[i + 1] + m) % m + (col[i + 1] - col[lst] + m) % m <= m)
		{
			++ans;
			lst = i + 1;
		}
	}
	printf("%d\n", ans);
	return 0;
}